J+1 maltiply top and bottom by (a+l) 40+44 40 G(O 2+1m2+1 4 4 ox又 ((o) 0+ O2+1)Vo2+1 o=tan@ Y=aGlo)sin(ot+9+ Y=2.6sinl 3t-1 2022-2-11 6
2022-2-11 6 G j j G j j j G j Y aG j t Y t ( ) ( ) ( ) tan ( ) sin . sin w w w w w w w w w w w w w w w w w 4 1 4 4 1 4 1 4 1 4 1 4 1 4 1 2 6 3 12 2 2 2 2 2 2 2 2 1 0 multiply top and bottom by (-j +1)
Bode Diagrams o a Bode diagram is a graphical method for illustrating how the phase angle and amplitude ratio of a particular system vary with frequency of the input sinusoid. For example consider the integrating system: G(s)=, the amplitude ratio and phase angle were determined above to be: AR= 90 The table below shows how these values vary with frequency AR 201og(ar) o 0.01 40dB 9 0.1 20dB 90° 1.0 OdB 90 0.1 -20dB 90° 100 0.01 40dB 2022-2-11 7
2022-2-11 7 Bode Diagrams ¨ A Bode diagram is a graphical method for illustrating how the phase angle and amplitude ratio of a particular system vary with frequency of the input sinusoid. For example consider the integrating system: , the amplitude ratio and phase angle were determined above to be: . The table below shows how these values vary with frequency: G s s ( ) 1 AR 1 90 0 w , AR 20log(AR) 0.01 100 40dB 0.1 10 20dB 1.0 1 0dB 10 0.1 -20dB 100 0.01 -40dB w 90 90 90 90 90
These values can now be plotted. a Bode diagram has two plots, one for the ar (in db) and the other for the phase angle Each plot has a log-linear scale. The Bode plot for this example Is shown below 111 I ;} 11 50 10 101 10 10 Frequency (rad/sec) """ a-90 1 144 "" t重1 10 10 10 10 10 Frequency〔rad/set 2022-2-11
2022-2-11 8 These values can now be plotted. A Bode diagram has two plots, one for the AR (in dB) and the other for the phase angle. Each plot has a log-linear scale. The Bode plot for this example is shown below:
Bode diagram of a First order System For a first order system the ar and phase angle can be determined as follows G(S) GGo 1+)0 GaindB=20 log 1+7ol 1-7)1-7c 1+⑦)o1+7o1-7o1+T 1+ta jo 1(1+702) VI+T Gain,p=2010g 11+10 Gains 20log√1+7 ∠=tan-oT 2022-2-11 9
2022-2-11 9 Bode Diagram of a First Order System ¨ For a first order system the AR and phase angle can be determined as follows: G s Ts G j Tj Gain Tj Tj Tj Tj Tj Tj T Tj T T T Gain T Gain T T dB dB dB ( ) ( ) log * log log tan 1 1 1 1 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 20 1 1 20 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 w w w w w w w w w w w w w w w w
For low frequencies, aT<l, therefore GaindB=0dB For high frequencies, oT>>1, therefore Gaind B=-20logoT Therefore, the gain plot can be represented by 2 asymptotes one line at odb and another line with -20dB/decade 0=1/T 20g1+7≈-3B O=2/T 20logV1+027≈6B O=10/7-20lg√1+02m2≈-20B For the phase angle tan ot 1+joT 0,p=00=1/7,=-45 →>∞φ>-90 2022-2-11 10
2022-2-11 10 For low frequencies, wT<<1, therefore GaindB=0dB For high frequencies, wT>>1, therefore GaindB=-20logwT Therefore, the gain plot can be represented by 2 asymptotes: one line at 0dB and another line with -20dB/decade. w w w w w w 1 20 1 3 2 20 1 6 10 20 1 20 2 2 2 2 2 2 / log / log / log T T dB T T dB T T dB For the phase angle w w w w w 1 1 0 0 1 45 90 1 j T T T o o o tan ; / ; ;